RI 2019 Year 5 CT Suggested Solutions

© Raffles Institution 2019 9729/J/19 2019 Y5 H2 Chemistry Term 3 Common Test – Suggested Solutions Section A Question 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Answer B C B B D C B C C D A A D A D MCQ worked solutions 1 (Ans: B) From the mass numbers given in the table, the sample contains 10B, 11B and 23Na (23Na is the only impurity present). To compute the Ar of boron in the given sample, we need to consider only the isotopes , 10B and 11B, and their % abundance (i.e. 23Na and i ts % abundance should not be included in the calculation). Note also that the sum of % abundance of 10B and 11B is not 100%. Ar of boron in sample = ( )( ) ( )( )15.52 11 74.48 15.52   10  +     + 74.48 = 10.83 2 (Ans: C) In the redox reaction, V2O5 is reduced and Fe2+ is oxidised to Fe3+. 25VOn : en − : 2+Fen Since Oxidation: Fe2+ ⎯→ Fe3+ + e− or 2Fe2+ ⎯→ 2Fe3+ + 2e− Reduction (unbalanced): V2O5 + 2e– ⎯→ ?  Each V atom accepted 1 e–.  Oxidation no. of V decreases from +5 to +4. : 1 : 1 1 : : 2 1 : 2 : 2 Oxidation number of V 0 +2 +4 +3 A V B VO C VO2 D V2O3 3 (Ans: B) H+, being positively charged, gives  of +15o and | q m| = 1 Since particle Z gives  of –5o, particle Z is negatively charged and | q m| = 1 3 proton neutron electron q m |q m| 1 1 0 4 –3 1 3 2 1 2 2 –1 3 1 3 3 3 3 1 +2 6 1 3 4 3 3 5 –2 6 1 3 Hence, only options 2 and 4 are correct. This document is copyrighted, please do not reproduce it without permission

© Raffles Institution 2019 9729/J/19 4 (Ans: B) There is a large jump in the 4th to 5th IE for one element, and in the 6th to 7th IE for another element. Hence these two elements belong to group 14 and 16 respectively, with minimally 8 electrons. CS2 is not possible because C has only 6 electrons to be removed. Hence only SiO2 is possible. 5 (Ans: D) A (incorrect): X and Y do not have the same proton number and hence they are not isotopes. B (incorrect): Electronegativity of an element increases across the period. Hence Y has a higher electronegativity than X. C (incorrect): Since X and Y have different proton number (and hence different no. of e−), they will not have the same electronic configuration. D (correct): As the ion of Y is isoelectronic with X+, the ion of Y has a +2 charge (i.e. Y2+). Hence, the ion of Y has a higher charge density (higher charge, smaller ionic radius) than the ion of X. 6 (Ans: C) molecule structure no. of  bonds no. of σ bonds CO2 2 2 HCN 2 2 N2 2 1 NO2 1 2 (Note: the dative covalent bond is also a σ bond) 7 (Ans: B) around underlined atom molecule structure no. of bond pairs no. of lone pairs shape BCl3 3 0 trigonal planar AlCl3 3 0 trigonal planar CO2 2 0 linear SiO2 4 0 tetrahedral H2O 2 2 bent SO2 2 1 bent Thi